The density of lead is `11.35 g cm^(-3)` and the metal crystallizes with fee unit cell. Estimate the radius of lead atom. (At. Mass of lead `= 207 g mol^(-1) and NA = 6.02xx10^23 mol^(-1))`

Answer

281 pm

Explanation

` P = ( Z xx M)/(a^(3) xx N_(A)) or a^(3) = ( 4 xx 107.87 " g mol"^(-1))/(( 406 xx 10^(-10) "cm")^(3) xx ( 6.02 xx 10^(23) "mol"^(-1) xx 11.35 "g cm" ^(-3))) = 12.118 xx 10^(-23) = 121.18 xx 10^(-24) `
or ` a = (121.18)^(1//3) xx 10^(-8)`
Let ` x = (121.18)^(1//3) therefore log x = 1/3 log 121.18 = 1/3 ( 2.083) = 0.694 or x = " antilog" 0.694 = 4.943`
a = 4.943 ` xx 10^(-8) " cm" = 493 .3 xx 10^(-10) " cm" = 494. 3 " pm" `
For fcc, ` r= a/(2sqrt2) = 0. 3535 a = 0.3535 xx 494 . 3 " pm" = 174.7 "pm" `